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What I want to prove : $$R_{ijrs}=\partial_r\Gamma_{ijs}-\partial_s\Gamma_{ijr}-\Gamma_{rj}^k\Gamma_{iks}+\Gamma_{sj}^k\Gamma_{ikr}$$ According to Riemann-Christoffel tensor, the covariant composant are given by: $$R_{ijrs}=g_{jk} R_{ijk}^k $$ We replace the following quantities $$g_{jk} \partial_r\Gamma_{is}^k$$ with: $$\partial_r (g_{jk}\Gamma_{is}^k) -\Gamma_{is}^k \partial_r g_{jk}$$ So, in the first identity we get the following: $$R_{ijrs}=\partial_r (g_{jk}\Gamma_{is}^k)-\partial_s (g_{jk}\Gamma_{ir}^k)+\Gamma_{is}^l(\Gamma_{rjl}-\partial_r g_{jl})-\Gamma_{ir}^l(\Gamma_{sjl}-\partial_s g_{jl})$$ And I'm stuck here :'( !!

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  • $\begingroup$ Consider making your question more clear. As it stands, it’s not clear what you’re asking. $\endgroup$ Mar 22, 2020 at 23:18
  • $\begingroup$ I'm asking for a proof, What I add was my try, I want to prove what I asked for :) !! @BobKnighton $\endgroup$
    – user257151
    Mar 22, 2020 at 23:20
  • $\begingroup$ @BobKnighton: it's pretty clear to me what the OP is asking - he's asking for help to prove the equation $R_{ijrs}=\partial_r\Gamma_{ijs}-\partial_s\Gamma_{ijr}-\Gamma_{rj}^k\Gamma_{iks}+\Gamma_{sj}^k\Gamma_{ikr}$ - but got lost in the math. $\endgroup$ Mar 24, 2020 at 3:42

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I'll give you hint.

Use $R$ and the coordinate vectors of the coordinate system

$$R(\partial_k,\partial_l)\partial_j=\nabla_{\partial_l}(\nabla_{\partial_k}\partial_j)-\nabla_{\partial_k}(\nabla_{\partial_l}\partial_j).$$

Note that

$$\nabla_{\partial_k}(\partial_j)=\Gamma^{m}_{kj}\partial_{m}.$$

Also, you only have to calculate the first term on the right of the equal sign - for the second term you just have to flip the necessary indices.

Then use the metric on $R$ to obtain the desired form if necessary.

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  • $\begingroup$ Thank you!! I'll try solving it using your hint, Thanks again $\endgroup$
    – user257151
    Mar 23, 2020 at 1:10
  • $\begingroup$ Also, to get to the final result, I took the inner product of the following equation $$R(\partial_k,\partial_l) \partial_j = \sum_{i} R^{i}_{jkl} \partial_{i}$$ with the tangent vector $ \partial_n$ $$\langle R(\partial_k,\partial_l) \partial_j,\partial_{n}\rangle = \sum_i R^{i}_{jkl}g_{in}=R_{jkln} $$ where $g_{in}=\langle \partial_i,\partial_n \rangle.$ Then I just re-mapped the indices on right side of the question you need to prove. $\endgroup$ Mar 24, 2020 at 3:40